# The need for phase wrapping?

I understand the need for phase unwrapping and the complexity behind it. But, I'm struggling to understand why the phase information gets wrapped between $-\pi$ to $\pi$ in the first place?

• What would it look like to have a complex number with a phase of $\frac{\pi}{2}$? What about a phase of $\frac{5\pi}{2}$? – Jason R Jan 11 '16 at 16:35
• Yes, it would be the same. But, for a finite number of different phases for a given frequency would it not be better to not wrap them? Where am I going wrong? – JJT Jan 11 '16 at 16:45

$$H(\omega)=|H(\omega)|e^{j\phi(\omega)}\tag{1}$$
where $\phi(\omega)$ is the phase, then you probably know that for a given frequency $\omega_0$, the frequency response might as well be written as
$$H(\omega_0)=|H(\omega_0)|e^{j(\phi(\omega_0)+2k\pi)}\tag{2}$$
where $k$ is some integer. The question is, when you compute the complex value $H(\omega_0)$, which of the infinitely many possible phases do you choose? The convention is to choose the principal value which is in the interval $(-\pi,\pi]$. The consequence is that whenever you compute the function $\phi(\omega)$ for continuous $\omega$, you'll get jumps as soon as the phase hits $-\pi$ or $\pi$, but no matter how you define the phase, you will not be able to avoid jumps. So if you want a continuous function you need to do phase unwrapping.